# Convergence of Taylor Series of Function Analytic on Disk/Lemma

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## Lemma

Let $y > 1$.

Then:

- $\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n} = 0$

## Proof

Note that $\ln y > 0$ as $y > 1$.

\(\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n}\) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \frac n {\left({e^{\ln y} }\right)^n}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \frac n {e^{\left({\ln y}\right) n} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | as $\displaystyle \lim_{x \mathop \to \infty} \frac x {e^{\left({\ln y}\right) x} } = 0$ by Limit at Infinity of Polynomial over Complex Exponential as $\ln y > 0$ |

$\blacksquare$